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Compute $\lim_{n \to \infty} \sum_{k=1}^{n} \arccos k$.
My attempt :the sequence is strictly increasing and if I prove that it is not upper bounded its limit is $\infty$. However, I can't prove the last part.

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  • $\begingroup$ Are you computing $\arccos 2$ with complex numbers? $\endgroup$ – J.G. Nov 14 '18 at 16:06
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    $\begingroup$ What is $\arccos(10)$? $\endgroup$ – ablmf Nov 14 '18 at 16:06
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$$\cos^{-1}(x)=\frac{\pi}{2}+i\log(\sqrt{1-x^2}+ix)$$ So $$\lim_{x \to \infty} \frac{1}{i}\cos^{-1}(x)=\infty$$

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  • $\begingroup$ Isn't arccos multivalued? $\endgroup$ – N. S. Nov 14 '18 at 16:11
  • $\begingroup$ @N.S. Is it? I think it's up to your definition. Of course, $\cos^{-1}$ is formally not a function but a relation, but you can define an "arccos" function. $\endgroup$ – Botond Nov 14 '18 at 16:15
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Well, $\arccos(k)$ is not defined for $k\ge 1$ when we consider it as a real function. So the question only makes sense when we consider it as a complex function. Thus $$ \cos ^{-1}(k)=\frac{\pi }{2}+i \log \left(\sqrt{1-k^2}+i k\right). $$ Then obviously $\sum_{k \ge 1} \cos ^{-1}(k)$ does not converge.

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